Question

b) In the adjacent figure, QR is the diameter of the circle with centre 'O' If
15 cm, PR = 8 cm, find the area of the shaded region. (Use a=3.14).
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Answer 1

Answer:

Area of a circle=πr²

Area of a semi-circle= πr²/2

In triangle POR,

PO=OR...... (radii of the same circle)

angle POR =90°

By Pythagoras theorem,

(PR)²=(OP)²+(OR)²

(8)²= 2(OP)²

64/2=(OP)²

32=OP²

Taking square roots,

OP= 4√2 cm

therefore radius of the circle=4√2cm

Area of a triangle=1/2 x base x height

triangle POR=1/2 x 4√2 x 4√2

= 16cm²

trianglePOQ=1/2 x 4√2 x4√2

=16cm²

Area of triangle PQR=16+16

=32cm²

area of semi circle=πr²/2

=3.14 x (4√2)²/2

=3.14 x 16 x 2/2

=3.14 x. 16

=18.84cm²

area of shaded region= area of semi circle. -area of triangle PQR

=32-18.84

=13.16cm²

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